In the book of Mathematical Analysis II by Zorich, at page 136, it is asked that
Let $f : I_{a,b} → R$ be a continuous function defined on an interval $I_{a,b} = \{x ∈ \mathbb{R}^n | a_i ≤ x_i ≤ b_i , i = 1, . . . , n\}$, and let $F : I_{a,b} → \mathbb{R}^1 $ be defined by the equality
$$F(x) = \int_{I_{a,x}} f(t) dt,$$
where $I_{a,x} ⊂ I_{a,b}$. Find the partial derivatives of this function with respect to the variables $x_1, . . . , x_n$.
Proof:
Directly from the definition of partial derivative,
$$\frac{\partial F}{\partial x_1 } (x) = \lim_{h \to 0} \frac{ \int_{I_{a, x'} } f(t)dt - \int_{I_{a,x}} f(t) dt}{ |h| },$$
where $x' = (x_1 + h, x_2, ..., x_n)$.
$$ = \lim_{h \to 0} \frac{ \int_{I_{a, x'} } f(t)dt - \int_{I_{a,x}} f(t) dt}{ |h| } $$
Lets define $G(x) = \int_{(a_2, a_3,..., a_n)}^{(x_2, x_3, ..., x_n)}$, then (since the double integral exist, we can deal with iterated integrals) $$= \lim_{h \to 0} \frac{ \int_{a_1}^{x_1 + h} G(t)dt - \int_{a_1}^{x_1} G(t)dt }{ |h| } = \lim_{h \to 0} \frac{ \int_{x_1}^{ x_1+ h} G(t)dt + \int_{a_1}^{x_1} G(t)dt - \int_{a_1}^{x_1} G(t)dt }{ |h| } \\ = \lim_{h \to 0} \frac{ \int_{x_1}^{x_1 + h} G(t)dt}{ |h| },$$
and by the Fundamental theorem of calculus, this equals to $$ = G(x)$$
Note: $G(x)$ denotes the $(n-1)$ dimensional integral of $f$ over the interval $I_{(a_2, a_3,..., a_n), (x_2, x_3, ..., x_n)}$.
Question:
Is there anything wrong with this proof ?